Boys' Toddler Uggs, AP® CALCULUS BC 2009 SCORING GUIDELINES. PDFA diver leaps from the edge of a diving platform into a pool below. The figure above shows the initial position of the diver and her position at a later.Questions & answersGauthmathQuestionA diver leaps from the edge of a diving platform into a pool below. The figure to the right shows the initial position of the diver and her position a a later time. At time t seconds after she leaps. the horizontal distance from the front edge of the platform to the diver's shoulders is given by xt, and the vertical distance from the water surface to her shoulders is given by vt, where xt and yt are measured in meters. Suppose that the diver's shoulders are 11.4 meters above the water when she makes her leap and that dx/dt =0.8 and dy/dt =3.6-9.8tfor 0 ≤ t ≤ A, where A is the time that the diver's shoulders enter thewater.Find the maximum vertical distance from the water surface and the diver's shouldersNote: Figure not drawn to scalAnswer · 69 votesAnswer: solve for t, 3.6t-9.8=0: t= 49/18 Decimal: t=2.72222. ∈ t 3.6t-9.8dt=1.8t2-9.8t+Cyt=1.8t2-9.8t+c.MoreBrainly.comQuestionA diver leaps from the edge of a diving platform into a pool below. The figure shows the initial position of the diver and her position at a later time. At time t seconds after she leaps, the horizontal distance from the front edge of the platform to the diver's shoulders is given by xXt and the vertical distance from the water surface to her shoulders is given by t), where x(t) and yt) are measured in meters. Suppose that the diver's shoulders are 11.4 meters above the water when she makes her leap and that d0.8 and dar 3.6 9.8t for Os ts A, where A is the time that the diver's shoulders enter the water. image y0) Note: Figure not drawn to scale Question 1 (3 points) Find the maximum vertical distance from the water surface to the diver's shoulders. Question 2 (2 points) Find A, the time that the diver's shoulders enter the water Question 3 (2 points) Find the total distance traveled by the diver's shoulders from the time she leaps from the platform until the time her shoulders enter the water. Question 4 (2 points) Find the angle e. 0くθ 〈 π , between the path of the diver and the water at the instant the diver's shoulders enter the water.Answer · 0 votesA- The maximum vertical distance from the water surface to the diver’sshoulders are 12.061 meters.B- The time that the diver’s shoulders enter the water is 1.936 seconds. C- The total distance traveled by the diver’s shoulders from the timeshe leaps from the platform until the time her shoulders enter the water is 12.946 meters.D- The angle, θ between the path of the diver and the water at the instant the diver’s shoulders enter the water is 1.519.A- [tex]frac{dy}{dx} =0[/tex] only when [tex]t=0.36735[/tex]. Let [tex]b=0.36735[/tex]The maximum vertical distance from the water surface to the diver’sshoulders is y(b)=11.4 + [tex]intlimits^b_0 frac{dy}{dT} dt = 12.061[/tex] meters.B- [tex]y(A)=11.4+intlimits^A_0 frac{dy}{dt} dt = 11.4+3.6A- 4.9A^{2} =0[/tex] when A= 1.936 seconds.C- [tex]intlimits^A_0 {sqrt{(frac{dx}{dt} )^{2} + (frac{dy}{dt} )^{2} } } , dt = 12.946 m[/tex]At time A, dy/dx = [tex]frac{dy/dt}{dx/dt} | _{t=A} = -19.21913[/tex]D- The angle between the path of t…MoreStudy.comQuestion. An 80 kg diver leaps from the end of 2.0-meter long diving board which has a mass of 10.0 kg.Answer · 0 votesPart a:In order to calculate the force on the supports, we have to analyze how the equilibrium is maintained. We build an equation for equilibrium analyzing how the forces exert torque on the board. We set the pivot at one of the 2 supports. We select the leftmost support.τnet=0mdgL+mbgL2−Fs2×0.3=080×9.8×2+10×9.8×1−Fs2×0.3=01568+98−0.3Fs2=0Fs2=16660.3=5553.3 NFinally to find the force exerted by the other support, build an equilibrium equation analyzing all of the forces working on the vertical axis.Fy=0Fs1+Fs2−mbg−mdg=0Fs1+5553.3−10×9.8−80×9.8=0Fs1+4671.33=0Fs1=−4671.33 NPart b:As the diver jumps, he exerts a force on the edge of the diving board equal to:F=mdaF=80×2F=160 NWe solve again with this new force on the problem.τnet=0FL+mbgL2−Fs2×0.3=0160×2+10×9.8×1−Fs2×0.3=0320+98−0.3Fs2=0Fs2=4180.3=1393.33 NAnd again, we solve by building an equilibrium equation working on the y-axis.Fy=0Fs1+Fs2−mbg−mdg=0Fs1+1393.33−10×9.8−160=0Fs1+1135.33=0Fs1=−1135.33 NMoreNumeradeQuestionAn Olympic diver leaps up off a diving board with a speed of 10 m/s at an angle of 50 degrees above the horizontal. The board is 250 meters above the pool: a) How long does it take her to hit the water? b) How far horizontally is she from the edge of the board when she hits the water? c) What is her velocity (both magnitude and direction!) just before she hits the water?Answer · 2 votesAnd this problem, we have been given that there is a swimmer and the swimmer is standing at a distance of 1.2 m above the surface of the water, and from this point the swimmer makes a jump with an initial velocity of five m/s. So we need to determine the total time that the swimmer's feet is in there. So for that we observed that the initial velocity is five m/s. The displacement here is -1.2 m because we take the downward direction as the negative direction, and acceleration also is acting in the downward direction, so it will be minus 9.8 meter per second square. So here we need to figure out the time. So for that we make use of the expression as equals U T plus half 80 square. So putting the values here, we get the quadratic equation That is -1.2 equals five T -4.90 square. So rearranging this quadratic equation, Wicked! 4.90 sq -50 -1.2 is equal to zero. So now we saw this quadratic equation to get the value of time too, Solving this wicket T as one two seconds approximately. So it…MoreNumeradeQuestionA diver stands on a diving platform 10.0 m above the surface of a pool and leaps upward with an initial speed of 2.5 m/s. how fast is the diver moving while falling past a diving board that is 3.0 m above the surface of the pool?Answer · 1 voteHello guys. And this problem we have to cover the concept of conservation of mechanical energy. Now from the conservation of mechanical energy. The initial mechanical energy of today's Evo will be Cuban into the final mechanical energy after dialogue and he hits the water. The initial mechanical energy will be in the form of gravitational potential energy and the kind of technology that is equals MG times H plus half times V. I square. And if you consider at the water level, the gravitational potential energy is zero. The final energy will be in the form of the kind of technology that is half times um V. F squared. Now we can cancel out the mask from each tell him of the equation and we got the initial speed equals squirreled off the final speech squared minus two G times the height of phone. Now substitute the values. Therefore the initial speed imposed Squire 8.2 m 4 2nd sq minus two times 9.8 m per second squared times the height a fall that is off three m are the initial speech of …More Feedback Nenul Clothing, CALCULUS BC - SECTION II, Part A - Time-45 minutes. 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